a block of mass m=0.9 kg is connected to a spring of force constant k= 775 N/m on a smooth horizontal surface. plot the potential energy of the spring from x= -5 cm to x= 5 cm. determine the turning points of the block if its speed at x= 0 is 1.2 m/s.
To plot the potential energy, simply plug your givens into the equation U=(1/2)kx^2. You should get a parabola that goes to 0 in the center.
To determine the turning points of the block, use the law of conservation of energy. You know that Ui+Ki = Uf+Kf since there is no friction. At x=0, there will be no potential energy (Ui=0). A "turning point" is where the motion will stop and the block will move in the opposite direction. This means that Kf=0.
So, set Ki=Uf
(1/2)mv^2 = (1/2)kx^2
(1/2)(.9)(1.2)^2 = (1/2)(775)x^2
Solve for x=.0409 m
This will be the first turning point. The other turning point will be the opposite of this (remember the parabola) - so x=-.0409 m.
To plot the potential energy of the spring from x = -5 cm to x = 5 cm, we need to use Hooke's Law, which states that the potential energy of a spring is given by:
PE = (1/2)kx^2,
where PE is the potential energy, k is the force constant of the spring, and x is the displacement from the equilibrium position.
Given:
Mass of the block (m) = 0.9 kg
Force constant of the spring (k) = 775 N/m
First, we need to convert the distance from centimeters to meters.
For x = -5 cm:
x = -5 cm = -0.05 m
For x = 5 cm:
x = 5 cm = 0.05 m
Now, we can calculate the potential energy at each value of x:
For x = -0.05 m:
PE = (1/2) * 775 * (-0.05)^2 = 0.484375 J (Joules)
For x = 0.05 m:
PE = (1/2) * 775 * (0.05)^2 = 0.484375 J (Joules)
As we can see, the potential energy is symmetric about the equilibrium position. Therefore, the graph of potential energy versus x will be a parabolic curve symmetrically distributed around the x-axis from x = -5 cm to x = 5 cm.
Now, let's determine the turning points of the block.
At a turning point, the block momentarily stops before changing direction. Therefore, the kinetic energy at a turning point is zero.
Given:
Velocity at x = 0 (v) = 1.2 m/s
To determine the turning points, we can use the equation:
KE = (1/2)mv^2,
where KE is the kinetic energy and m is the mass of the block.
Since KE = 0 at a turning point, we can solve for v:
(1/2) * m * v^2 = 0
Since mass (m) and velocity (v) are positive, the only possibility for KE to be zero is when v = 0.
Therefore, the turning points of the block occur when the velocity is zero, which means the block is momentarily at rest.
In this case, the turning points will occur at x = -5 cm and x = 5 cm, where the potential energy is at its maximum value.